Last updated at Dec. 16, 2024 by Teachoo
Misc 6 Prove that ((sinβ‘γ7π₯ + sinβ‘γ5π₯) + (sinβ‘γ9π₯ + sinβ‘γ3π₯)γ γ γ γ)/((cosβ‘γ7π₯ + πππ 5π₯) + (cosβ‘γ9π₯ + cosβ‘γ3π₯)γ γ γ ) = tan 6x Solving L.H.S ((sinβ‘γ7π₯ + sinβ‘γ5π₯) + (sinβ‘γ9π₯ + sinβ‘γ3π₯)γ γ γ γ)/((cosβ‘γ7π₯ + πππ 5π₯) + (cosβ‘γ9π₯ + cosβ‘γ3π₯)γ γ γ ) Lets solve numerator and Denominator separately Solving numerator (sin 7x + sin 5x) + ( sin 9x + sin 3x) = 2 sin ((7π₯ + 5π₯)/2) cos ((7π₯ β 5π₯)/2) + 2sin ((9π₯ +3π₯)/2) cos ((9π₯ β3π₯)/2) = 2 sin (12π₯/2) . cos (2π₯/2) + 2sin (12π₯/2) cos (6π₯/2) = 2 sin 6x . cos x + 2 sin 6x . cos 3x = 2 sin 6x (cos x + cos 3x) Now solving Denominator (cos 7x + cos 5x) + (cos 9x + cos 3x) = 2 cos ((7π₯ + 5π₯)/2) cos ((7π₯ β 5π₯)/2) + 2 cos ((9π₯ +3π₯)/2) cos ((9π₯ β3π₯)/2) = 2 cos (12π₯/2) . cos (2π₯/2) + 2 cos (12π₯/2) cos (6π₯/2) = 2 cos 6x . cos x + 2 cos 6x . cos 3x = 2 cos 6x (cos x + cos 3x) Now solving L.H.S ((sinβ‘γ7π₯ + sinβ‘γ5π₯) + (sinβ‘γ9π₯ + sinβ‘γ3π₯)γ γ γ γ)/((cosβ‘γ7π₯ + πππ 5π₯) + (cosβ‘γ9π₯ + cosβ‘γ3π₯)γ γ γ ) = (π πππβ‘ππ (πππβ‘π + πππβ‘ππ))/(π πππβ‘ππ (πππβ‘π + πππβ‘ππ)) = (sinβ‘6π₯ )/(cosβ‘6π₯ ) = tan 6x = R.H.S Hence L.H.S = R.H.S Hence proved
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo